Work,energy COM and linear momentum conservation

In summary: Better choose the ground as the reference frame. In summary, the conversation discusses a problem involving a wagon with a pendulum attached to it on a frictionless surface. The velocity of the wagon when the pendulum reaches a certain angle is to be determined. The solution involves three equations, one relating the horizontal components of the velocities of the wagon and the pendulum, another relating the total energy of the system, and a third involving the angle y and the velocities. The reference frame of the ground is recommended for simplicity.
  • #1
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Homework Statement


A wagon with mass M can move on frictionless surface. A mathematical/ideal pendulum is fastened on the wagon. At the initial moment the wagon and the pendulum were at rest and the pendulum makes an angle of x with the vertical. What will be the velocity of the wagon when the pendulum makes an angle of y with the vertical?


Homework Equations


I guess its a linear momentum question but I don't know the equation.


The Attempt at a Solution

 
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  • #2
Hi,

1 - You nearly got the 1st thing right. Since friction is neglected, the system wagon-pendulum experiences no horizontal force, and thus its horizontal component of linear momentum is conserved. From this, you have the 1st equation relating the horizontal components of the velocities of the wagon and the bob [tex]v_{h(wagon)}[/tex] and [tex]v_{h(bob)}[/tex] respectively.

2 - The 2nd conserved thing is the total energy of the system, which comprises of the potential energy of the bob and the kinetic energies of the wagon and the bob. This is the 2nd equation, which relates [tex]v_{h(wagon)}[/tex] , [tex]v_{h(bob)}[/tex] , [tex]v_{v(bob)}[/tex] - the vertical component of the bob's velocity and the angles x, y.

3 - The 3rd condition is that in the reference frame of the wagon, the velocity of the bob is perpendicular to the string. You should deduce the 3rd equation relating [tex]v_{h(wagon)}[/tex] , [tex]v_{h(bob)}[/tex] , [tex]v_{v(bob)}[/tex] and the angle y.

Try to write down the equations :smile:
 
  • #3
well i m not sure of the first equation.
is it mv(bob)= - Mv(wagon)??
 
  • #4
It's [tex]mv_{h(bob)} = - Mv_{h(wagon)}[/tex]. (horizontal component!)
 
  • #5
but the velocity of bob is it relative to the ground??
 
  • #6
Yes. If you choose the reference frame of the wagon from the start, you have to take into account the fictitious force, which is unnecessarily complicated.
 

1. What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, but can only be transferred or transformed from one form to another. This means that the total energy in a closed system remains constant over time.

2. How does the conservation of energy relate to work?

The conservation of energy is related to work in the sense that work is a transfer of energy from one object to another. When work is done on an object, energy is transferred to that object, and when work is done by an object, energy is transferred away from that object.

3. What is the center of mass (COM) and how is it conserved?

The center of mass is the point at which the mass of a system is evenly distributed. It is conserved because the total momentum of a system is equal to the mass of the system multiplied by the velocity of the center of mass. This means that as long as there are no external forces acting on the system, the center of mass will remain constant.

4. How is linear momentum conserved?

Linear momentum is conserved in a closed system, meaning that the total momentum of all objects in the system remains constant. This is because, according to Newton's third law of motion, for every action, there is an equal and opposite reaction. This means that any changes in momentum of one object will be matched by an equal and opposite change in momentum of another object in the system.

5. Can energy and linear momentum be transferred between objects?

Yes, energy and linear momentum can be transferred between objects through various interactions such as collisions or forces. However, the total energy and momentum in a closed system will remain constant.

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